Problem: The altitude to the hypotenuse of a triangle with angles of 30 and 60 degrees is 3 units. What is the area of the triangle, in square units? Express your answer in simplest radical form. [asy]
unitsize(6mm);
defaultpen(linewidth(.7pt)+fontsize(8pt));

real r=2*sqrt(3);

pair A=r*dir(0), B=r*dir(60), C=r*dir(180);
pair F=foot(B,A,C);
draw(A--B--C--cycle);
draw(rightanglemark(A,B,C,8));
draw(B--F,linetype("4 2"));
label("3 units",waypoint(B--F,0.6),W);
[/asy]
Solution: Let $A$, $B$, and $C$ be the vertices of the right triangle corresponding to the 60, 90, and 30 degree angles, respectively.  Also, let $F$ be the foot of the perpendicular from $B$ to the hypotenuse $AC$.  Notice that $\bigtriangleup BAF$ is a 30-60-90 triangle.  The longer leg of a 30-60-90 is $\sqrt{3}$ times the shorter leg, so $AF=3/\sqrt{3}=\sqrt{3}$ units.  The hypotenuse of a 30-60-90 triangle is twice the shorter leg, so $AB=2\sqrt{3}$ units.  Since $\bigtriangleup CAB$ is a 30-60-90 triangle as well, $BC=2\sqrt{3}\cdot\sqrt{3}=6$ units.  The area of $ABC$ is $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(2\sqrt{3})(6)=\boxed{6\sqrt{3}}$ square units.

[asy]
unitsize(6mm);
defaultpen(linewidth(.7pt)+fontsize(8pt));
real r=2*sqrt(3);
pair A=r*dir(0), B=r*dir(60), C=r*dir(180);
pair F=foot(B,A,C);
draw(A--B--C--cycle);
draw(rightanglemark(A,B,C,8));
draw(B--F,linetype("4 2"));
label("3",waypoint(B--F,0.6),W);

label("$A$",A,SE);
label("$B$",B,N);
label("$C$",C,SW);
label("$F$",F,S);[/asy]